Chapter 3 - Worked Solutions

to selected Complex Reasoning questions from the OUP text Senior Physics - Knowledge, Processes and Reasoning by Walding, Rapkins and Rossiter

Question 30

(a)

(b)

(c)

BC = 8 sin 18° = 2.5 ms-1

BD = 5.0 ms-1 in direction W18° S

 

Question 31

(a) FH = 100 sin 60° = 87 N

(b) FV = 100 cos 60° = 50N

 

Question 32

vPG = 250/0.5 = 500 km-1 E

vAG = 50 kmh-1 N (not S); hence vGA = 50 kmh-1 S

vPA = vPG + vGA = 500 kmh-1 E + 50 kmh-1 S

 

 

vPA = 502.5 kmh-1 ; angle q = tan-1 50/500 = 5.7° ; hence direction is E 5.7° S

 

Question 33

dP = 4 dQ and rP = 4 rQ

(a) AP = 4p rP2

AQ = 4p (rP/4)2 = 4p rP2 /16 = AQ/16

Hence AP = AQ/16

(b) VP = 4/3 p rP3

VQ = 4/3 p rQ (1/4)3

Hence VP = VQ/64

(c) mP = r VP

mQ = r VQ

mP = mQ/64

(d) r P = r Q

 

Question 34

(a) V1 = 35 V2

4/3 p r13 = 35 x 4/3 p r22

(b)



Question 37 (New Century edition)

Square both sides of the equation to give: Z2 = R2 + 4p 2L2f2
Rearrange the equation to: Z2 = 4p 2L2f2 + R2
This is now in the form y = mx + c
where the slope (m) = 4p 2L2 and the intercept on the y-axis (c) = R2.
Plot the graph:

Slope = (44100 - 10000)/90000 = 0.379
4p 2L2 = 0.379
L = 0.098 metres

Intercept = 10000 = R2
Therefore R = 100 ohm

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